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As you sit in a fishing boat, you notice that 12 waves pass the boat every 45 s
. If the distance from one crest to the next is 9.0 m
, what is the speed of these waves?
Express your answer to two significant figures and include the appropriate units.

The system's final temperature is roughly 16.7°C.

You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.

The following is the formula for energy conservation:

Q1 + Q2 = 0

Q = mcΔT

Q1 + Q2 = 0

568.8

Simplifying and solving for

6394.4 - 106768 = 0

= 16.7°C

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The original temperature of the mercury is 260.6K

To solve this problem, we can use the formula for the heat released during the solidification of a substance:

Q = m * Lf

where Q is the heat released, m is the mass of the substance, and Lf is the heat of fusion of the substance.

In this case, Q = 157 kJ, m = 5.00 kg, and Lf = 11.3 kJ/kg.

We also need to use the formula for the heat absorbed or released during a temperature change:

Q = m * c * ΔT

where Q is the heat absorbed or released, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature.

We can use this formula to calculate the heat released as the mercury cools from its original temperature to its melting point, and then use the formula for solidification to calculate the heat released as the mercury solidifies.

Let T be the original temperature of the mercury.

The heat released as the mercury cools from its original temperature to its melting point is:

Q1 = m * c * (T - 234)

The heat released as the mercury solidifies is:

Q2 = m * Lf

The total heat released is:

Q = Q1 + Q2 = m * c * (T - 234) + m * Lf

Substituting the values given in the problem, we get:

157 kJ = 5.00 kg * 140 J/kg . K * (T - 234) + 5.00 kg * 11.3 kJ/kg

Simplifying and solving for T, we get:

T = 260.6 K

Therefore, the original temperature of the mercury was 260.6 K.

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The above has to do with the study of the earth's lithospheric plates. See the attached image and the explanation below.

The movement of lithospheric plates is a geological process that occurs due to the motion of hot, molten material in the Earth's mantle. The lithosphere, which is the rigid outer layer of the Earth's surface, is divided into several large plates that move relative to each other.

These movements are caused by the convection of material in the mantle and the forces that arise at the boundaries between the plates.

There are three main types of plate boundaries: divergent, convergent, and transform. Divergent boundaries occur where plates move apart from each other, creating new oceanic crust. Convergent boundaries arise where plates collide, leading to subduction, volcanic activity, and the formation of mountains. Transform boundaries occur where plates slide past each other.

The movement of lithospheric plates gives rise to various geological phenomena, such as earthquakes, volcanic activity, and the formation of mountain ranges and ocean basins.

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Answer: d. increase

Explanation:

If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.

The light ray will deviate from its original path with 19.5° after refraction.

Applying Snell's law to calculate the angle of refraction:

n1 sin θ1 = n2 sin θ2

where n1 and θ1 =  the refractive index and the angle of incidence in the first medium (air),

n2 and θ2 =  the refractive index and the angle of refraction in the second medium (glass).

In this example,

n1 = 1.00 (refractive index of air), θ1 = 30°, and

n2 = 1.5 (refractive index of glass).

We then calculate for  θ2:

n1 sin θ1 = n2 sin θ2

1.00 * sin 30° = 1.5 * sin θ2

0.5 = 1.5 * sin θ2

sin θ2 = 0.5 / 1.5 = 1/3

θ2 = sin^-1(1/3)

θ2 = 19.5°

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Answer:

Explanation:

Because you have velocity along the y axis and time along the x axis, this is a velocity v time graph which is an acceleration graph. The slope of the line in this graph IS the acceleration. We can use 2 points and the slope formula to solve for the acceleration:

(0, 0) and (1, 3):

[tex]m=\frac{3-0}{1-0}=3[/tex] m/s squared, choice D.

The value of the charge on each particle is [tex]1.05 x 10^-8 C[/tex].

Coulomb's law is a fundamental principle of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can use Coulomb's law to solve this problem. Mathematically,

[tex]F = k(q1q2)/r^2[/tex]

where F is the force of attraction or repulsion between the two charged particles,[tex]q1[/tex] and [tex]q2[/tex] are the magnitudes of the charges on the two particles, r is the distance between them, and k is Coulomb's constant, which has a value of [tex]9.0 x 10^9 Nm^2/C^2.[/tex]

In this problem, we know that the charges are equal and the distance between them is 10.0 cm. We also know that the force between them is 0.5 N. Therefore,

[tex]0.5 N = k(q^2)/(0.1 m)^2[/tex]

Solving for q, we get:

[tex]q = \sqrt{[(0.5 N)(0.1 m)^2/k]}[/tex]

[tex]q = \sqrt{(0.5 N)(0.01 m)/(9.0 x 10^9 Nm^2/C^2)}[/tex]

[tex]q = 1.05 x 10^-8 C[/tex]

Therefore, the value of the charge on each particle is [tex]1.05 x 10^-8 C.[/tex]

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a) The time after the release of the first stone that the second stone hits the water is 2.0 s.

b) 15.7 m/s is the initial speed of the second stone.

c)  The speed of the first stone as it hits the water is 15.7 m/s.

d) The speed of the second stone as it hits the water is 28.2 m/s.

Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).

a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.

Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.

b) The initial speed of the second stone can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (1.7)² + 2(9.8) * 59

v = 15.7 m/s

c) The speed of the first stone as it hits the water can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (1.7)² + 2(9.8) * 59

v = 15.7 m/s

d) The speed of the second stone as it hits the water can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (15.7)² + 2(9.8) * 59

v = 28.2 m/s

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The electric force between two point charges is given by the equation

[tex]F=k*q_1*q_2/r^2[/tex]

The interaction between two things is measured by the physical quantity known as force. It is a vector quantity, and the sign F is frequently used to denote it. When an object interacts with another object, it feels a push or a pull.

where r is the distance between the charges, q1 and q2 are their magnitudes, and k is the Coulomb constant.

When we enter the problem's specified values, we obtain

[tex]2.2N=8.99*10^9\ N*m^2/C^2*q*-2q/(1.4 m)^2[/tex]

which simplifies to

q = -0.500 N/C.

Thus, the magnitude of charge q is 0.500 N/C.

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Answer:

B. Atomic number minus mass number

Explanation:

The wavelength of the electron is 1.724 × 10^-12 m.

The wavelength of the electron is found  using the de Broglie wavelength formula:

λ = h / p

where λ = wavelength,

h= Planck's constant, a

p =  momentum of the electron.

we find  the momentum of the electron,

p = m * v

p = (9.1 × 10^-31 kg) * (4.2 × 10^7 m/s)

p = 3.822 × 10^-22 kg m/s

Therefore, wavelength ;

λ = h / p

λ = (6.6 × 10^-34 J s) / (3.822 × 10^-22 kg m/s)

λ = 1.724 × 10^-12 m

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A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).

9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.

The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.

The following formula may be used to determine how long 1.5 million radiation cycles take to complete:

Time is equal to the frequency of cycles.

Plugging in the numbers, we get:

time = 1.5 million / 9.192631770 × 10^9 Hz

time = 1.632995101 × 10^-7 seconds

So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.

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A sound that is 7.8x10-4 W/m2 in intensity is equal to (10 dB)log3.2106 W/m21012 W/m2=185 dB.

The decibel, often known as the db or 0.1 bel, is the standard measurement unit. Hence, b = 10 log10 (I/I0) can be used to express the relationship between relative intensities, or b, in decibels. This equation can be used to determine that one decibel equals a 26 percent intensity variations.

The "decibel level" of a sound is a less formal term for relative intensity level. It is not the same as energy; relative intensity level reflects loudness more faithfully by using a logarithmic scale.

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